@jamestanton God wants us to write our polynomials like this:
1ax^0
1ax^1 + 1bx^0
1ax^2 + 2bx^1 + 1cx^0
1ax^3 + 3bx^2 + 3cx^1 + 1dx^0
1ax^4 + 4bx^3 + 6cx^2 + 4dx^1 + 1ex^0,
with their coefficients multiplied by entries drawn from Pascal's triangle. Makes everything nicer.
This computation leads to the so-called "depressed cubic" which is a cubic polynomial in ax + b with no quadratic term. And one can continue in this same way to arrive at "depressed" quartics, quintics, etc. They're all polynomials in ax + b, with no subleading term.
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And since all these "depressed" polynomials are polynomials in ax + b, we know that when the coefficient sequence a, b, c, ... is geometric, they must take the form (ax + b)^n = 0 -- which means that their coefficients must all vanish in this case. Like ac - b^2, ad - bc, etc.
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Indeed, one easily finds, for example, that a^2(ax^3 + 3bx^2 + 3cx + d) = (ax + b)^3 + 3(ac - b^2)(ax + b) + a(ad - bc) - 2b(ac - b^2). And this "depressed" cubic in ax + b has for its coefficients polynomials in ac - b^2 and ad - bc -- both zero in the geometric case.
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I thought you might enjoy this string of ideas, which one can easily make into a "general theory of depression." It goes farther than this, too, as I imagine will not surprise you. Really quite lovely stuff, which puts "completing the square" into its proper algebraic context.
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A final remark: in this context, the cubic formula takes on a peculiarly elegant and memorable form: if ax^3 + 3bx^2 + 3cx + d = 0, then ax + b = cbrt[(ac - b^2)(ar + b)] + cbrt[(ac - b^2)(as + b)], where r, s are the roots of (ac - b^2)x^2 + (ad - bc)x + (bd - c^2) = 0.
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End of conversation
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